'Reciprocity' in $\mathbb{Z}[\zeta_n]$

112 Views Asked by At

Let $K=\mathbb{Q}(\zeta_n)$ where $\zeta_n$ is an $n^{th}$ root of unity, let $\mathfrak{p}$ be a prime ideal in $K$ over the rational prime $p$, and let $\alpha\in \mathbb{Z}[\zeta_n]$. Then is there a way to determine weather or not there exists an $\eta\in\mathbb{Z}[\zeta_n]$ such that $\eta^p\equiv \alpha \pmod{ \mathfrak{p}}$?

More generally, if $\mathfrak{a}$ is an arbitrary (integral) ideal in $K$ and $t$ is a positive integer, then is there a way of determining wether or not there is a solution to $\eta^t\equiv \alpha\pmod{\mathfrak{a}}$ for some $\eta\in\mathbb{Z}[\zeta_n]$?

(It seems that the power residue symbol gives an answer when $t=n$.)

2

There are 2 best solutions below

0
On

The quotient ring $k=\Bbb Z[\zeta_n]/\mathfrak{p}$ is a finite field of characteristic $p$. Within $k$ the map $\eta\mapsto\eta^p$ is an automorphism (the Frobenius automorphism). So the answer to whether there is an $\eta\in\Bbb Z[\zeta_n]$ such that $\eta^p\equiv\alpha\pmod{\mathfrak p}$ is "always".

5
On

$\eta \mapsto \eta^p$ is the Frobenius automorphism of $\mathbb{Z}[\zeta_n]/\mathfrak{p}$ and $\alpha^{p^f} \equiv \alpha\bmod \mathfrak{p}$ so $\eta \equiv \alpha^{p^{f-1}}\bmod \mathfrak{p}$ where $f = [\mathbb{Z}[\zeta_n]/\mathfrak{p}:\mathbf{F}_p]=[\mathbf{F}_p(\zeta_n):\mathbf{F}_p] = \text{ord}(p \bmod n)$.

That there is a solution for $\eta^t \equiv \alpha\bmod \mathfrak{b}$ depends of course on $t$ and $\text{ord}(\alpha \bmod \mathfrak{b})$.